$$\begin{array}{cc} & |\overline{AC}|\,|\overline{FD}| = |AEDF| = |\overline{AB}|\,|\overline{ED}| \\[6pt] \to \qquad & \large{b^2\,\left(\,\cot A + \cot B\,\right) = c^2\,\left(\,\cot A + \cot C\,\right)} \\ \end{array}$$

## Archive for the ‘**Trigonography**’ Category

## Cotangent Identity

## Half-Angles in a Triangle

My answer to this question at Math.StackExchange.com. $$\Large{c\;\sin\frac{A-B}{2} \;\equiv\; (a-b)\;\cos\frac{C}{2}}$$

## Identity from Math.StackExchange.com

Taken from my answer to “Proving a weird trig identity” at Math.StackExchange.com . $$\Large{\frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta + \tan\theta}{1}}$$ In the diagram, \(\overline{AB}\) is tangent to the unit circle at P. The “trig lengths” (except for \(|\overline{AQ}|\)) should be clear. We note that \(\angle BPR \cong \angle RPP^\prime\), since these inscribed angles subtend congruent arcs \(\stackrel{\frown}{PR}\) […]

## Product Identities

$$\begin{align} 2 \cos A \cos B &= \cos(A-B) \;+\; \cos(A+B) \\ 2 \sin A \;\sin B &= \cos(A-B) \;-\; \cos(A+B) \end{align}$$

## Proof Without Words: Nine-Point Circle Property

The “Nine-Point Circle” of a triangle \(\triangle ABC\) contains the vertices of the medial triangle (\(\triangle RST \)) and orthic triangle (\(\triangle UVW \)), as well as points \(X\), \(Y\), \(Z\) that bisect segments from \(\triangle ABC\)’s orthocenter (\(H\)) to its vertices. A question on Math.StackExchange.comĀ asked for proof that the center of the nine-point circle […]